# Commutative unitary ring without maximal ideal without axiom of choice

I want to find a semi-constructive example of a unitary commutative ring without any maximal ideals assuming that axiom of choice is incorrect and/or a model of $$\sf ZF$$ where we have such a concrete ring.

This question is similar to the questions Vector space bases without axiom of choice and A confusion about Axiom of Choice and existence of maximal ideals..

What I tried is to use $$\mathbb{R}$$ as a $$\mathbb{Q}$$ vector space without a basis and try to construct some chains of ideals on a related ring and try to show that a maximal ideal corresponds to a basis but didn't achieve much.

• Well. You can either construct special models with particular examples, and that normally involves techniques like forcing and stuff; or you can sort of repeat the proof that the existence of maximal ideals implies choice and start with a set that cannot be well-ordered as a counterexample. This set can be $\Bbb R$ in models where there is no Hamel basis, for example. Jul 7, 2019 at 14:46
Let $$k$$ be a field, let $$I\subset k^{\mathbb{N}}$$ be the ideal of sequences that are eventually zero, and let $$S=k^{\mathbb{N}}/I$$. Then maximal ideals in $$S$$ are in bijection with nonprincipal ultrafilters on $$\mathbb{N}$$. In particular, in any model of ZF in which there are no nonprincipal ultrafilters on $$\mathbb{N}$$, there will be no maximal ideals in $$S$$.